var
f:array[0..2000,0..2000]of longint;
n,kk,i,j,k,ans:longint;
begin
readln(n,kk);
for i:=1 to n do f[1,i]:=1;//1是每个数的倍数
for i:=1 to kk-1 do
for j:=1 to n do//n个数
for k:=1 to n div j do//n是n div j的倍数
f[i+1,j*k]:=(f[i+1,j*k]+f[i,j])mod 1000000007;//取模
for i:=1 to n do ans:=(ans+f[kk,i])mod 1000000007;
writeln(ans);
end.